MATH 220 (all sections)—Homework #12 not to be turned in posted Friday, November 24, 2017 Definition: A set A is finite if there exists a nonnegative integer c such that there exists a bijection from A to fn 2 N: n ≤ cg. (The integer c is called the cardinality of A.) I. (a) Let A be a finite set, and let B be a subset of A. Prove that B is finite. (Hint: induction on jAj. Note that our proof can’t use induction on jBj, or indeed refer to “the number of elements in B” at all, because we don’t yet know that B is finite!) (b) Prove that the union of two disjoint finite sets is finite. (c) Prove that the union of any two finite sets is finite. (Hint: A [ B = A [ (B − A).) (a) We proceed by induction on the nonnegative integer c in the definition that A is finite (the cardinality of c). Basis step: c = 0. Then there is a bijection from A to fn 2 N: n ≤ 0g = ;, and thus A = ; (and, for that matter, the bijection is also the empty function). This forces B = ; as well (since that is the only subset of ;), which is certainly finite. Induction step: Let c ≥ 0. The induction hypothesis is: if A has cardinality c, then any subset of A is finite. Now suppose A has cardinality c + 1, and let f : A ! fn 2 −1 N: n ≤ c + 1g be a bijection. Define A1 = f (f1; : : : ; cg), and note that f restricted to A1 is a bijection from A1 to f1; : : : ; cg, so that A1 has cardinality c. Finally, let B be a subset of A. −1 −1 Case 1: f (c + 1) 2= B. Then B is a subset of f (f1; : : : ; cg) = A1, which has cardinality c; by the induction hypothesis, B is finite. −1 −1 Case 2: f (c + 1) 2 B. Then define B1 = B − f (c + 1). As before, B1 is a subset of A1 and hence finite. Let d be the cardinality of B1 and let g1 : B1 ! fn 2 N: n ≤ dg be a bijection. Now define a function g : B ! fn 2 N: n ≤ d + 1g by ( g (x); if x 2 B ; g(x) = 1 1 d + 1; if x = f −1(c + 1): It is not hard to check that g is a bijection from B to fn 2 N: n ≤ d + 1g. In particular, B is finite. (b) Let C and D be finite sets with C \ D = ;, and let f : C ! fn 2 N: n ≤ cg and g : D ! fn 2 N: n ≤ dg be bijections. Define h: C [ D ! fn 2 N: n ≤ c + dg by ( f(x); if x 2 C; h(x) = g(x) + c; if x 2 D: It is (somewhat tedious but) not hard to check that h is a bijection from C [ D to fn 2 N: n ≤ c + dg. (The fact that C \ D = ; is necessary to show that h is a well-defined function.) In particular, C [ D is finite. (c) Let A and B be finite sets. Then B − A is a subset of the finite set B and hence is itself finite by part (a). Consequently, since A and B − A are always disjoint, the set A [ B = A [ (B − A) is the union of two disjoint finite sets and is therefore finite by part (b). II. Let A and B be nonempty sets. Prove that there exists an injective function f : A ! B if and only if there exists a surjective function g : B ! A. First assume that f : A ! B is injective. Let D = f(A) be the range of A; then f is a bijection from A to D. Choose any a 2 A (possible since A is nonempty). Define g : B ! A by ( f −1(y); if y 2 D; g(y) = a; if y 2 B − D: It is not hard to show that g is a well-defined function that is surjective. Conversely, assume that g : B ! A is surjective. Define a function f : A ! B as follows: for each a 2 A, choose ba 2 B such that g(ba) = a, and define f(a) = ba. It is not hard to check that f is a well-defined function that is injective. [Interested students who just read this proof might wish to find some introductory information about something called the “Axiom of Choice”.] SECTION 13.1 Show that the two given sets have equal cardinality by describing a bijection from one to the other. Describep your bijection with a formula (not as a table). 2. R and ( 2; 1) 4. The set of even integers and the set of odd integers 8. Z and S = fx 2 R: sin x = 1g 10. f0; 1g × N and Z 14. N × N and f(n; m) 2 N × N: n ≤ mg. (Hint: draw “graphs” of both sets. The northwest–southeast diagonal slices of the first set look a lot like the horizontal slices of the second set. .) p 2. Adaptingp one of the bijections from the book, define f : R ! ( 2; 1) by f(x) = x e + 2. It’s easyp to check that f is a well-defined and bijective function (for example, f −1(y) = ln(y − 2) is its inverse function). 4. A bijection f : feven integersg ! fodd integersg is f(n) = n + 1; or f(n) = n − 1; or indeed f(n) = n + k or f(n) = n − k for any fixed odd integer k. 7π 3π π 5π 8. Note that S = fx 2 R: sin x = 1g = f:::; − 2 ; − 2 ; 2 ; 2 ;::: g. We see that one π bijection f : Z ! S is f(n) = 2πn + 2 . 10. Define a function f : f0; 1g × N ! Z by ( n; if b = 0; f(b; n) = 1 − n; if b = 1: It is not hard to show that f is a bijection; for example, its inverse function is ( (0; m); if m ≥ 1; f −1(m) = (1; 1 − m); if m ≤ 0: 14. Inspired by the given hint, define a function f : f(n; m) 2 N × N: n ≤ mg ! N × N by f(n; m) = (n; m + 1 − n). Since n ≤ m for elements of the domain, we see that m+1−n ≥ 1, and so the values really do lie in the codomain. One can check that this function is a bijection; for example, its inverse function is f −1(x; y) = (x; x+y−1). III. Suppose that two sets A and B have the same cardinality. Prove that P(A) and P(B) have the same cardinality as each other. Let f : A ! B be a bijection. Define a function g : P(A) !P(B) by g(X) = f(X) for any X 2 P(A). (Let’s interpret the notation carefully: g(X) is a function value since X is an element of the domain P(A) of g, while f(X) is an image since X is a subset of the domain A of f.) We claim that g is a bijection. First, let Y 2 P(B). Then if we set X = f −1(Y ), then g(X) = f(X) = f(f −1(Y )) = Y . (The identity f(f −1(Y )) = Y isn’t true in general, but it is true when f is surjective, as shown on the previous homework.) In particular, g is surjective. Next, let X1 and X2 be elements of P(A) and suppose that g(X1) = g(X2). Then −1 −1 f(X1) = f(X2) (as images). Taking preimages of both sets yields f (f(X1)) = f (f(X2)). −1 −1 Since f is injective, we have f (f(X1)) = X1 and f (f(X2)) = X2 as shown on the previous homework; therefore X1 = X2. In particular, h is surjective. SECTION 13.2 2. Prove that the set A = f(m; n) 2 N × N: m ≤ ng is countably infinite. We already know that N×N is countably infinite. By problem 14 in Section 13.1 (above), A has the same cardinality as N × N. Therefore A is also countably infinite. 4. Prove that the set of all irrational numbers is uncountable. Let I be the set of irrational numbers. Suppose for the sake of contradiction that I is countably infinite. We already know that Q is countably infinite. Then R = Q[I, the union of two countably infinite sets, would be countably infinite as well; but this contradicts the known fact that R is uncountable. Therefore I is not countably infinite; since I is certainly an infinite set, we conclude that I is uncountable. 6. Prove or disprove: There exists a bijective function f : Q ! R. Disproof: if there were such a bijective function, then Q and R would have the same cardinality. But we know that Q is countably infinite while R is uncountable, and therefore they do not have the same cardinality. We conclude that there is no bijection from Q to R. 8. Prove or disprove: The set Z × Q is countably infinite. Proof: we know that both Z and Q are countably infinite, and we know that the Cartesian product of two countably infinite sets is again countably infinite. Therefore Z × Q is countably infinite. 12. Describe a partition of N that divides N into @0 countably infinite subsets. We know that N × N is countably infinite, so let f : N × N ! N be a bijection. For each m 2 N, define Am = f fmg × N .